Find $\left \lceil \frac{12}{7} \cdot \frac{-29}{3}\right\rceil - \left\lfloor \frac{12}{7} \cdot \left \lfloor \frac{-29}{3}\right \rfloor \right \rfloor$.
Solution: Evaluating the first term, $\frac {12}7 \cdot \frac{-29}{3} = \frac{-116}{7}$. Since $$-17 = \frac{-119}{7} < \frac{-116}{7} < \frac{-112}{7} = -16,$$ the ceiling of $\frac{-116}{7}$ is $-16$.

In the second term, since $$-10 = \frac{-30}{3} < \frac{-29}{3} < \frac{-27}{3} = -9,$$ then the floor of $\frac{-29}3$ is $-10$. The product of this with $\frac{12}{7}$ is $\frac{-120}{7}$. Since $$-18 = \frac{-126}{7} < \frac{-120}{7} < \frac{-119}{7} = -17,$$ the floor of $\frac{-120}{7}$ is $-18$. Hence, the answer is $-16 - (-18) = \boxed{2}$.